Problem: What is the average value of $\cos(x)$ on the interval $[-2,7]$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\sin(7)+\sin(2)}{9}$ (Choice B) B $\dfrac{\sin(7)-\sin(2)}{9}$ (Choice C) C $\dfrac{\sin(7)+\sin(2)}{5}$ (Choice D) D $\dfrac{\sin(7)-\sin(2)}{5}$
Solution: In general, this is the average value of function $f$ over the interval $[a,b]$ : $\dfrac{\int_a^b f(x)\,dx}{b-a}$ In our case, ${f(x)=\cos(x)}$, ${a=-2}$ and ${b=7}$ : $\begin{aligned} \dfrac{\int_{ a}^{ b} {f(x)}\,dx}{ b- a}&=\dfrac{\int_{{-2}}^{ 7} {\cos(x)}\,dx}{ 7-({-2})} \\\\ &=\dfrac{\Big[\sin(x)\Big]_{-2}^7}{9} \\\\ &=\dfrac{\sin(7)-\sin(-2)}{9} \\\\ &=\dfrac{\sin(7)+\sin(2)}{9} \end{aligned}$ In conclusion, this is the average value of $\cos(x)$ on the interval $[-2,7]$ : $\dfrac{\sin(7)+\sin(2)}{9}$